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| 1. |
Find k so that the point P(-4,6) lies on the line segment joining A(k,10) and B(3,-8). |
| Answer» If P (-4, 6) lies on the line segment joining A (k, 10) and B (3, -8), then P, A and B are collinear.{tex} \\therefore{/tex}(-4 {tex} \\times{/tex}\xa010 + k {tex} \\times{/tex}\xa0-8 + 3 {tex} \\times{/tex}\xa06) - (6k + 30 + -4 {tex} \\times{/tex}\xa0-8) = 0{tex} \\Rightarrow{/tex}(-40 - 8k + 18) - (6k + 30 + 32) = 0{tex} \\Rightarrow{/tex}(-22 - 8k) - (6k + 62) = 0{tex} \\Rightarrow{/tex}\xa0-14k - 84 = 0{tex} \\Rightarrow{/tex}\xa0k = - 6Suppose P divides AB in the ratio {tex} \\lambda{/tex}: 1. Then, the coordinates of Pare {tex} \\left( \\frac { 3 \\lambda - 6 } { \\lambda + 1 } , \\frac { - 8 \\lambda + 10 } { \\lambda + 1 } \\right){/tex}.But, the coordinates of P are (-4, 6).{tex} \\therefore \\quad \\frac { 3 \\lambda - 6 } { \\lambda + 1 } = - 4 \\text { and } \\frac { - 8 \\lambda + 10 } { \\lambda + 1 } = 6{/tex}{tex} \\Rightarrow \\quad \\lambda = \\frac { 2 } { 7 }{/tex}Hence, P divides AB in the ratio\xa0{tex} \\frac 27{/tex} : 1 or 2: 7. | |